Find the probability of :
(i) getting 5 exactly twice in 7 throws of a die
(ii) throwing at most 2 sixes in 6 throws of a single die.

Answer

To solve the given problems, we will use the concepts of binomial probability.

### Part (i): Finding the probability of getting 5 exactly twice in 7 throws of a die.

1. **Identify the parameters:**
- Number of trials (n) = 7 (the number of throws)
- Number of successes (r) = 2 (we want to get a 5 exactly twice)
- Probability of success (p) = 1/6 (the probability of rolling a 5)
- Probability of failure (q) = 1 - p = 5/6 (the probability of not rolling a 5)

2. **Use the binomial probability formula:**
The binomial probability formula is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
Substituting the values:
\[ P(X = 2) = \binom{7}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{7-2} \]

3. **Calculate the binomial coefficient:**
\[ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21 \]

4. **Substitute and calculate:**
\[ P(X = 2) = 21 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^5 \]
\[ = 21 \times \frac{1}{36} \times \left(\frac{5^5}{6^5}\right) \]
\[ = 21 \times \frac{1}{36} \times \frac{3125}{7776} \]
\[ = \frac{21 \times 3125}{36 \times 7776} \]
\[ = \frac{65625}{279936} \]

5. **Final probability:**
\[ P(X = 2) \approx 0.234 \]

### Part (ii): Finding the probability of throwing at most 2 sixes in 6 throws of a single die.

1. **Identify the parameters:**
- Number of trials (n) = 6
- Probability of success (p) = 1/6 (the probability of rolling a 6)
- Probability of failure (q) = 5/6

2. **Calculate the probability of getting 0, 1, or 2 sixes:**
We need to find \( P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \).

- **For \( P(X = 0) \):**
\[ P(X = 0) = \binom{6}{0} \left(\frac{1}{6}\right)^0 \left(\frac{5}{6}\right)^6 = 1 \times 1 \times \left(\frac{5}{6}\right)^6 = \left(\frac{5}{6}\right)^6 \]

- **For \( P(X = 1) \):**
\[ P(X = 1) = \binom{6}{1} \left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^5 = 6 \times \frac{1}{6} \times \left(\frac{5}{6}\right)^5 = 6 \times \frac{5^5}{6^6} \]

- **For \( P(X = 2) \):**
\[ P(X = 2) = \binom{6}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^4 = 15 \times \left(\frac{1}{6}\right)^2 \times \left(\frac{5}{6}\right)^4 \]

3. **Combine the probabilities:**
\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]
\[ = \left(\frac{5}{6}\right)^6 + 6 \times \frac{5^5}{6^6} + 15 \times \frac{5^4}{6^6} \]

4. **Factor out \( \frac{1}{6^6} \):**
\[ = \frac{1}{6^6} \left(5^6 + 6 \times 5^5 + 15 \times 5^4\right) \]

5. **Calculate the final probability:**
\[ = \frac{1}{6^6} \left(15625 + 6 \times 3125 + 15 \times 625\right) \]
\[ = \frac{1}{6^6} \left(15625 + 18750 + 9375\right) \]
\[ = \frac{1}{6^6} \times 43750 \]
\[ \approx 0.651 \]

### Final Answers:
(i) The probability of getting 5 exactly twice in 7 throws of a die is approximately \( 0.234 \).

(ii) The probability of throwing at most 2 sixes in 6 throws of a single die is approximately \( 0.651 \).